Question: Add the following rational expressions. $\dfrac{5x}{2x-3}+\dfrac{-4x^2}{3x+1}=$
We can add two rational expressions whose denominators are equal by adding the numerators and keeping the denominator the same. [Does this fit with how we add rational numbers?] When the denominators are not the same, we must manipulate them so that they become the same. In other words, we must find a common denominator. Since the two denominators do not share any common factors, the common denominator is simply the product of these two denominators: $({2x-3})\cdot({3x+1})$. Let's manipulate the expressions to have that denominator: $\begin{aligned} &\phantom{=}\dfrac{5x}{{2x-3}}+\dfrac{-4x^2}{{3x+1}} \\\\ &=\dfrac{5x\cdot({3x+1})}{({2x-3})\cdot({3x+1})}+\dfrac{-4x^2\cdot({2x-3})}{({3x+1})\cdot({2x-3})} \end{aligned}$ [Why did we do that?] Now that both denominators are the same, let's add! $\begin{aligned} &\phantom{=}\dfrac{5x\cdot(3x+1)}{(2x-3)\cdot(3x+1)}+\dfrac{-4x^2\cdot(2x-3)}{(3x+1)\cdot\!(2x-3)} \\\\ &=\dfrac{5x\cdot(3x+1)+(-4x^2)\cdot\!(2x-3)}{(2x-3)(3x+1)} \\\\ &=\dfrac{15x^2+5x-8x^3+12x^2}{(2x-3)(3x+1)} \\\\ &=\dfrac{-8x^3+27x^2+5x}{(2x-3)(3x+1)} \end{aligned}$ In conclusion, $\dfrac{5x}{2x-3}+\dfrac{-4x^2}{3x+1}=\dfrac{-8x^3+27x^2+5x}{(2x-3)(3x+1)}$